Solve the following questions. (Geometry: Similarity)
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Consider similar triangles ABE and BCD, then we have ∠BDC=∠EAB.
For triangle ABE,
tan(∠EAB)=EBAB
For triangle BCD,
tan(∠BDC)=BCBD
Thus we have
EBAB=BCBD⇒EB⋅BD=AB⋅BC
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Answer:
Answer
i) In △DGH & △DFE
∠D=∠D (Common)
∠DHG=∠DEF (Corresponding angle)
∠DGH=∠DFE (Corresponding angle)
∴△DGH∼△DFE (by AAA)
ii) If angles are a° then △ABC &△DEF are similar to each other.
iii) △POQ & △MON
OM
OP
=
6+2
6
=
8
6
=
4
3
ON
OQ
=
8+2
8
=
10
8
=
5
4
MN
PQ
=
6
4
=
3
2
∴ All ratio are not equal. So, triangles are not similar.
iv) In △XUV & △XYZ
XY
XU
=
2a+a
a
=
3a
a
=
3
1
XZ
XV
=
2c+c
c
=
3c
c
=
3
1
YZ
UV
=
3b
b
=
3
1
∴ All ratio's are equal. So, given △XUV & △XYZ are similar to each
other.
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